Answer:
The solution to the given systems of linear equation is {3, -1} lies in IV Quadrant.
Step-by-step explanation:
Given:
x-3y = 6 ........( 1 )
x+y=2 ........( 2 )
To find :
x = ?
y = ?
Solution:
Let solve the above equation to get the solutions.
Eliminate X by subtracting the two equations equation 1 minus equation 2
[tex](x-3y) - (x+y) = 6 - 2\\-4y = 4\\\therefore y=-1[/tex]
Now substitute for y = -1 in equation 2 we get
[tex]x+(-1)=2\\\therefore x = 2+1\\\therefore x = 3[/tex]
Therefore, solution set is {3, -1}
In
I Quadrant {x,y} = {+,+}......both the coordinates are positive.
II Quadrant {x,y} = {-,+}.....X coordinate is negative and Y coordinate is positive.
III Quadrant {x,y} = {-,-}......both the coordinates are negative.
IV Quadrant {x,y} = {+,-}.....X coordinate is positive and Y coordinate is negative. {3,-1}
Point A on the graph is the solution.
